Let $A = (0,1]$. Then $\inf(A) = 0$

Question

I am having a problem with this statement. I am trying to prove that 0 is the greatest lower bound by showing that every lower bound greater than 0 is a contradiction but I can't figure our how.

Proposition.

Let $A = (0,1]$. Then $\inf(A) = 0$

Proof.

Here, $0 < a$ for all $$a \in A$$

So $0$ is a lower bound for $A$.

Now suppose there is some $$b \in R$$ such that b is a lower bound for $A$ and $b>0$

Then, assuming $b≤1$, since $$b \in A$$ b can not be a lower bound (is this part wrong? if right, why?)

Hence $0$ is the greatest lower bound and $\inf(A)$.

Answer 1

You're right. If $b \in A$ then $b$ cannnot be a lower bound, since, for exemple, $\frac{b}{2} \in A$ and $\frac{b}{2} < b$.

Answer 2

What properties does $\inf A$ have? Is the biggest number such that $$\inf A \le a $$ for every $a \in A$.

Now it is true that $0 \le a$ for all $a \in A$, as you say; so $\inf A \ge 0$.

Suppose now $\inf A = b > 0$; but then $\frac b2 \in A$ and $\frac b2 < b = \inf A$ contradicting the property that $\inf A$ is smaller than any element of $A$.

So $\inf A$ must be $\ge 0$ but cannot be $> 0$ hence it must be $\inf A = 0$